Problem: Divide the following complex numbers. $ \dfrac{-4+36i}{-4-4i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-4+4i}$ $ \dfrac{-4+36i}{-4-4i} = \dfrac{-4+36i}{-4-4i} \cdot \dfrac{{-4+4i}}{{-4+4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-4+36i) \cdot (-4+4i)} {(-4-4i) \cdot (-4+4i)} = \dfrac{(-4+36i) \cdot (-4+4i)} {(-4)^2 - (-4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-4+36i) \cdot (-4+4i)} {(-4)^2 - (-4i)^2} = $ $ \dfrac{(-4+36i) \cdot (-4+4i)} {16 + 16} = $ $ \dfrac{(-4+36i) \cdot (-4+4i)} {32} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-4+36i}) \cdot ({-4+4i})} {32} = $ $ \dfrac{{-4} \cdot {(-4)} + {36} \cdot {(-4) i} + {-4} \cdot {4 i} + {36} \cdot {4 i^2}} {32} $ Evaluate each product of two numbers. $ \dfrac{16 - 144i - 16i + 144 i^2} {32} $ Finally, simplify the fraction. $ \dfrac{16 - 144i - 16i - 144} {32} = \dfrac{-128 - 160i} {32} = -4-5i $